Central-Difference Formulas

Transcription

1 SEC 61 APPROXIMATING THE DERIVATIVE 323 Central-Difference Formulas If the function f (x) can be evaluated at values that lie to the left and right of x, then the best two-point formula will involve abscissas that are chosen symmetrically on both sides of x Theorem 61 (Centered Formula of Order O(h 2 )) that x h, x, x + h [a, b] Then (3) f f (x + h) f (x h) (x) 2h Furthermore, there exists a number c = c(x) [a, b] such that (4) f f (x + h) f (x h) (x) = + E trunc ( f, h), 2h where E trunc ( f, h) = h2 f (3) (c) = O(h 2 ) 6 The term E( f, h) is called the truncation error Assume that f C 3 [a, b] and Proof Start with the second-degree Taylor expansions f (x) = P 2 (x) + E 2 (x), about x,for f (x + h) and f (x h): (5) f (x + h) = f (x) + f (x)h + f (2) (x)h 2 and (6) f (x h) = f (x) f (x)h + f (2) (x)h 2 2! After (6) is subtracted from (5), the result is 2! + f (3) (c 1 )h 3 f (3) (c 2 )h 3 (7) f (x + h) f (x h) = 2 f (x)h + (( f (3) (c 1 ) + f (3) (c 2 ))h 3

2 324 CHAP 6 NUMERICAL DIFFERENTIATION Since f (3) (x) is continuous, the intermediate value theorem can be used to find a value c so that f (3) (c 1 ) + f (3) (c 2 ) (8) = f (3) (c) 2 This can be substituted into (7) and the terms rearranged to yield (9) f (x) = f (x + h) f (x h) 2h f (3) (c)h 2 The first term on the right side of (9) is the central-difference formula (3), the second term is the truncation error, and the proof is complete Suppose that the value of the third derivative f (3) (c) does not change too rapidly; then the truncation error in (4) goes to zero in the same manner as h 2, which is expressed by using the notation O(h 2 ) When computer calculations are used, it is not desirable to choose h too small For this reason it is useful to have a formula for approximating f (x) that has a truncation error term of the order O(h 4 ) Theorem 62 (Centered Formula of Order O(h 4 )) that x 2h, x h, x, x + h, x + 2h [a, b] Then Assume that f C 5 [a, b] and (10) f (x) f (x + 2h) + 8 f (x + h) 8 f (x h) + f (x 2h) 12h Furthermore, there exists a number c = c(x) [a, b] such that (11) f (x) = f (x + 2h) + 8 f (x + h) 8 f (x h) + f (x 2h) 12h + E trunc ( f, h), where E trunc ( f, h) = h4 f (5) (c) = O(h 4 ) 30 Proof One way to derive formula (10) is as follows Start with the difference between the fourth-degree Taylor expansions f (x) = P 4 (x) + E 4 (x), about x,of f (x + h) and f (x h): (12) f (x + h) f (x h) = 2 f (x)h + 2 f (3) (x)h f (5) (c 1 )h 5 5! Then use the step size 2h, instead of h, and write down the following approximation: (13) f (x + 2h) f (x 2h) = 4 f (x)h + 16 f (3) (x)h f (5) (c 2 )h 5 5!

3 SEC 61 APPROXIMATING THE DERIVATIVE 325 Next multiply the terms in equation (12) by 8 and subtract (13) from it The terms involving f (3) (x) will be eliminated and we get f (x + 2h) + 8 f (x + h) 8 f (x h) + f (x 2h) (14) = 12 f (x)h + (16 f (5) (c 1 ) 64 f (5) (c 2 ))h If f (5) (x) has one sign and if its magnitude does not change rapidly, we can find a value c that lies in [x 2h, x + 2h] so that (15) 16 f (5) (c 1 ) 64 f (5) (c 2 ) = 48 f (5) (c) After (15) is substituted into (14) and the result is solved for f (x), we obtain (16) f f (x + 2h) + 8 f (x + h) 8 f (x h) + f (x 2h) (x) = + f (5) (c)h 4 12h 30 The first term on the right side of (16) is the central-difference formula (10) and the second term is the truncation error; the theorem is proved Suppose that f (5) (c) is bounded for c [a, b]; then the truncation error in (11) goes to zero in the same manner as h 4, which is expressed with the notation O(h 4 ) Now we can make a comparison of the two formulas (3) and (10) Suppose that f (x) has five continuous derivatives and that f (3) (c) and f (5) (c) are about the same Then the truncation error for the fourth-order formula (10) is O(h 4 ) and will go to zero faster than the truncation error O(h 2 ) for the second-order formula (3) This permits the use of a larger step size Example 62 Let f (x) = cos(x) (a) Use formulas (3) and (10) with step sizes h = 01, 001, 0001, and 00001, and calculate approximations for f (08) Carry nine decimal places in all the calculations (b) Compare with the true value f (08) = sin(08) (a) Using formula (3) with h = 001, we get f f (081) f (079) (08) Using formula (10) with h = 001, we get f f (082) + 8 f (081) 8 f (079) + f (078) (08) ( ) 8( ) (b) The error in approximation for formulas (3) and (10) turns out to be and , respectively In this example, formula (10) gives a better approximation to f (08) than formula (3) when h = 001 The error analysis will illuminate this example and show why this happened The other calculations are summarized in Table 62

4 SEC 62 NUMERICAL DIFFERENTIATION FORMULAS Numerical Differentiation Formulas More Central-Difference Formulas The formulas for f (x 0 ) in the preceding section required that the function can be computed at abscissas that lie on both sides of x, and they were referred to as centraldifference formulas Taylor series can be used to obtain central-difference formulas for the higher derivatives The popular choices are those of order O(h 2 ) and O(h 4 ) and are given in Tables 63 and 64 In these tables we use the convention that f k = f (x 0 +kh) for k = 3, 2, 1, 0, 1, 2, 3 For illustration, we will derive the formula for f (x) of order O(h 2 ) in Table 63 Start with the Taylor expansions (1) f (x + h) = f (x) + hf (x) + h2 f (x) 2 + h3 f (3) (x) 6 + h4 f (4) (x) 24 + Table 63 Central-Difference Formulas of Order O(h 2 ) f (x 0 ) f 1 f 1 2h f (x 0 ) f 1 2 f 0 + f 1 h 2 f (3) (x 0 ) f 2 2 f f 1 f 2 2h 3 f (4) (x 0 ) f 2 4 f f 0 4 f 1 + f 2 h 4 Table 64 Central-Difference Formulas of Order O(h 4 ) f (x 0 ) f f 1 8 f 1 + f 2 12h f (x 0 ) f f 1 30 f f 1 f 2 12h 2 f (3) (x 0 ) f f 2 13 f f 1 8 f 2 + f 3 8h 3 f (4) (x 0 ) f f 2 39 f f 0 39 f f 2 f 3 6h 4

5 340 CHAP 6 NUMERICAL DIFFERENTIATION and (2) f (x h) = f (x) hf (x) + h2 f (x) 2 h3 f (3) (x) 6 + h4 f (4) (x) 24 Adding equations (1) and (2) will eliminate the terms involving the odd derivatives f (x), f (3) (x), f (5) (x), : (3) f (x + h) + f (x h) = 2 f (x) + 2h2 f (x) 2 Solving equation (3) for f (x) yields + 2h4 f (4) (x) 24 + (4) f (x) = f (x + h) 2 f (x) + f (x h) h 2 2h4 f (6) (x) 6! 2h2k 2 f (2k) (x) (2k)! 2h2 f (4) (x) 4! If the series in (4) is truncated at the fourth derivative, there exists a value c that lies in [x h, x + h], so that (5) f (x 0 ) = f 1 2 f 0 + f 1 h 2 h2 f (4) (c) 12 This gives us the desired formula for approximating f (x): (6) f (x 0 ) f 1 2 f 0 + f 1 h 2 Example 64 Let f (x) = cos(x) (a) Use formula (6) with h = 01, 001, and 0001 and find approximations to f (08) Carry nine decimal places in all calculations (b) Compare with the true value f (08) = cos(08) (a) The calculation for h = 001 is f f (081) 2 f (080) + f (079) (08) ( ) (b) The error in this approximation is The other calculations are summarized in Table 65 The error analysis will illuminate this example and show why h = 001 was best

6 SEC 62 NUMERICAL DIFFERENTIATION FORMULAS 341 Table 65 Numerical Approximations to f (x) for Example 64 Step Approximation by Error using size formula (6) formula (6) h = h = h = Error Analysis Let f k = y k + e k, where e k is the error in computing f (x k ), including noise in measurement and round-off error Then formula (6) can be written (7) f (x 0 ) = y 1 2y 0 + y 1 h 2 + E( f, h) The error term E(h, f ) for the numerical derivative (7) will have a part due to roundoff error and a part due to truncation error: (8) E( f, h) = e 1 2e 0 + e 1 h 2 h2 f (4) (c) 12 If it is assumed that each error e k is of the magnitude ɛ, with signs that accumulate errors, and that f (4) (x) M, then we get the following error bound: (9) E( f, h) 4ɛ h 2 + Mh2 12 If h is small, then the contribution 4ɛ/h 2 due to round-off error is large When h is large, the contribution Mh 2 /12 is large The optimal step size will minimize the quantity (10) g(h) = 4ɛ h 2 + Mh2 12 Setting g (h) = 0 results in 8ɛ/h 3 + Mh/6 = 0, which yields the equation h 4 = 48ɛ/M, from which we obtain the optimal value: (11) h = ( ) 48ɛ 1/4 M When formula (11) is applied to Example 64, use the bound f (4) (x) cos(x) 1 = M and the value ɛ = The optimal step size is h = ( /1) 1/4 = , and we see that h = 001 was closest to the optimal value

7 342 CHAP 6 NUMERICAL DIFFERENTIATION Since the portion of the error due to round off is inversely proportional to the square of h, this term grows when h gets small This is sometimes referred to as the step-size dilemma One partial solution to this problem is to use a formula of higher order so that a larger value of h will produce the desired accuracy The formula for f (x 0 ) of order O(h 4 ) in Table 64 is (12) f (x 0 ) = f f 1 30 f f 1 f 2 12h 2 + E( f, h) The error term for (12) has the form (13) E( f, h) = 16ɛ 3h 2 + h4 f (6) (c), 90 where c lies in the interval [x 2h, x + 2h] A bound for E( f, h) is (14) E( f, h) 16ɛ 3h 2 + h4 M 90, where f (6) (x) M The optimal value for h is given by the formula ( ) 240ɛ 1/6 (15) h = M Example 65 Let f (x) = cos(x) (a) Use formula (12) with h = 10, 01, and 001 and find approximations to f (08) Carry nine decimal places in all the calculations (b) Compare with the true value f (08) = cos(08) (c) Determine the optimal step size (a) The calculation for h = 01 is f (08) f (10) + 16 f (09) 30 f (08) + 16 f (07) f (06) (b) The error in this approximation is The other calculations are summarized in Table 66 (c) When formula (15) is applied, we can use the bound f (6) (x) cos(x) 1 = M and the value ɛ = These values give the optimal step size h = ( /1) 1/6 =

8 Numerical Methods Using Matlab, 4 th Edition, 2004 John H Mathews and Kurtis K Fink ISBN: Prentice-Hall Inc Upper Saddle River, New Jersey, USA